Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 76 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


Problem

Find the set of all values of {m} such that {\displaystyle {y} = {\frac{x^2-x}{1-mx}}} can take all real values.


Solution

{\displaystyle {y} = {\frac{x^2-x}{1-mx}}}
{\Leftrightarrow} {\displaystyle{y - myx = x^2 - x}}
{\Leftrightarrow} {x^2 + x(my - 1) - y = 0}
{\Leftrightarrow} {\displaystyle{x} = {\frac{1 - my {\pm} {\sqrt{(my-1)^2+4y}}}{2}}}
Now {y} takes all real values if discriminant {(my-1)^2 + 4y} is allways {> 0}.
So now we have to find the all values of {m} such that {(my-1)^2 + 4y} {> 0} for all {y} {\in} |R.
{\Leftrightarrow} {m^2 y^2 - 2my + 1 + 4y > 0}
{\Leftrightarrow} {m^2 y^2 + y(4 - 2m) + 1 > 0} …(i)
Now this is a equation of upside open parabola. If the discriminant is {\le{0}} of equation (i) then {(my - 1)^2 + 4y} will always positive.
{\Rightarrow} {\displaystyle{(4 - 2m)^2 - 4m^2} {\le {0}}}
{\Rightarrow} {{16m + 16} {\le {0}}}
{\Rightarrow} {m + 1 {le{0}}}
{\Rightarrow} {m {\le{- 1}}}

Conclusion:

If {m {\le{1}}}, then {\displaystyle{y = {\frac{x^2 - x}{1 - mx}}}} can take all the values as {x} varies over |R.